Practice PSet Discussion

Problem P-1: Basics

Throughout, let the (bosonic) Fock basis be

\[ \ket{\mathbf n}\equiv \ket{n_1,n_2,\dots} \]

with orthonormality \(\braket{\mathbf m}{\mathbf n}=\delta_{\mathbf m,\mathbf n}\).

We are given the defining action of the creation operator:

\[ \hat a_j^\dagger \ket{\mathbf n} = \sqrt{n_j+1}\,\ket{\mathbf n+\mathbf e_j}, \]

where \(\mathbf e_j\) is the occupation-number vector with a \(1\) in slot \(j\) and zeros elsewhere.

We will also use the corresponding annihilation action (consistent with \(\hat a_j = (\hat a_j^\dagger)^\dagger\)):

\[ \hat a_j \ket{\mathbf n} = \sqrt{n_j}\,\ket{\mathbf n-\mathbf e_j}, \qquad \hat a_j \ket{\dots,n_j=0,\dots}=0. \]

(a) Show that \([\hat a_i,\hat a_j]=0\)

Act on an arbitrary basis state \(\ket{\mathbf n}\).

Case 1: \(i\neq j\).

First apply \(\hat a_j\) then \(\hat a_i\):

\[ \hat a_i \hat a_j \ket{\mathbf n} = \hat a_i \Big(\sqrt{n_j}\,\ket{\mathbf n-\mathbf e_j}\Big) = \sqrt{n_j}\,\sqrt{n_i}\,\ket{\mathbf n-\mathbf e_j-\mathbf e_i}. \]

Reverse the order:

\[ \hat a_j \hat a_i \ket{\mathbf n} = \hat a_j \Big(\sqrt{n_i}\,\ket{\mathbf n-\mathbf e_i}\Big) = \sqrt{n_i}\,\sqrt{n_j}\,\ket{\mathbf n-\mathbf e_i-\mathbf e_j}. \]

The final states are identical, and the prefactors match, hence

\[ (\hat a_i\hat a_j-\hat a_j\hat a_i)\ket{\mathbf n}=0 \qquad (i\neq j). \]

Case 2: \(i=j\). Then \([\hat a_i,\hat a_i]=0\) trivially.

Since the commutator annihilates every basis vector, we conclude

\[ [\hat a_i,\hat a_j]=0. \]

(b) Show that \([\hat a_i,\hat a_j^\dagger]=\delta_{ij}\)

Again act on \(\ket{\mathbf n}\).

Case 1: \(i\neq j\).

Compute \(\hat a_i \hat a_j^\dagger \ket{\mathbf n}\):

\[ \hat a_i \hat a_j^\dagger \ket{\mathbf n} = \hat a_i\Big(\sqrt{n_j+1}\,\ket{\mathbf n+\mathbf e_j}\Big) = \sqrt{n_j+1}\,\sqrt{n_i}\,\ket{\mathbf n+\mathbf e_j-\mathbf e_i}. \]

Compute \(\hat a_j^\dagger \hat a_i \ket{\mathbf n}\):

\[ \hat a_j^\dagger \hat a_i \ket{\mathbf n} = \hat a_j^\dagger\Big(\sqrt{n_i}\,\ket{\mathbf n-\mathbf e_i}\Big) = \sqrt{n_i}\,\sqrt{n_j+1}\,\ket{\mathbf n-\mathbf e_i+\mathbf e_j}. \]

Same state, same prefactor, so

\[ (\hat a_i\hat a_j^\dagger-\hat a_j^\dagger \hat a_i)\ket{\mathbf n}=0 \qquad (i\neq j). \]

Case 2: \(i=j\).

First,

\[ \hat a_j \hat a_j^\dagger \ket{\mathbf n} = \hat a_j\Big(\sqrt{n_j+1}\,\ket{\mathbf n+\mathbf e_j}\Big) = \sqrt{n_j+1}\,\sqrt{n_j+1}\,\ket{\mathbf n} = (n_j+1)\ket{\mathbf n}. \]

Second,

\[ \hat a_j^\dagger \hat a_j \ket{\mathbf n} = \hat a_j^\dagger\Big(\sqrt{n_j}\,\ket{\mathbf n-\mathbf e_j}\Big) = \sqrt{n_j}\,\sqrt{n_j}\,\ket{\mathbf n} = n_j\ket{\mathbf n}. \]

Subtracting gives

\[ [\hat a_j,\hat a_j^\dagger]\ket{\mathbf n}=\ket{\mathbf n}. \]

Combining the two cases, for all \(i,j\):

\[ [\hat a_i,\hat a_j^\dagger]\ket{\mathbf n}=\delta_{ij}\ket{\mathbf n}. \]

Since this holds on a complete basis, we conclude the operator identity

\[ [\hat a_i,\hat a_j^\dagger]=\delta_{ij}. \]

(c) Show that \(\hat a_j^\dagger \hat a_j\) is the number operator for orbital \(j\)

Define

\[ \hat n_j \equiv \hat a_j^\dagger \hat a_j. \]

Act on \(\ket{\mathbf n}\):

\[ \hat n_j \ket{\mathbf n} = \hat a_j^\dagger\Big(\sqrt{n_j}\,\ket{\mathbf n-\mathbf e_j}\Big) = \sqrt{n_j}\,\sqrt{n_j}\,\ket{\mathbf n} = n_j \ket{\mathbf n}. \]

So every Fock state \(\ket{\mathbf n}\) is an eigenstate of \(\hat n_j\) with eigenvalue \(n_j\), i.e. \(\hat n_j\) returns “how many bosons occupy orbital \(j\).”

(Equivalently, the total number operator is \(\hat N=\sum_j \hat n_j\), with \(\hat N\ket{\mathbf n}=(\sum_j n_j)\ket{\mathbf n}\).)

Problem P-2: Bosons in a harmonic oscillator

Let the (single-particle) harmonic-oscillator energy eigenstates be \(\{\phi_j(x)\}_{j=1}^\infty\), with single-particle energies \(\{E_j\}_{j=1}^\infty\), so that \(\hat h\,\phi_j = E_j \phi_j\). For the harmonic oscillator, \(\omega=\sqrt{k/m}\) and (with the convention \(j=1,2,\dots\) corresponding to \(n=0,1,\dots\)) one has \(E_j=\hbar\omega\left(j-\tfrac{1}{2}\right)\).

The given (unnormalized) many-body state is

\[ \ket{\tilde\psi}=(\hat a_1^\dagger)(\hat a_2^\dagger)^3\ket{0}. \]

(a) Normalization factor using second quantization

We compute \(\braket{\tilde\psi}{\tilde\psi}\) and then divide by its square root.

Start from

\[ \braket{\tilde\psi}{\tilde\psi} = \bra{0}\hat a_2^3 \hat a_1\,\hat a_1^\dagger (\hat a_2^\dagger)^3\ket{0}. \]

Use the facts that different modes commute, \([\hat a_1,\hat a_2^\dagger]=0\), and also \(\hat a_1\hat a_1^\dagger = 1+\hat a_1^\dagger \hat a_1\) so that acting on the vacuum gives \(\hat a_1\hat a_1^\dagger\ket{0}=\ket{0}\). Then the expectation value factorizes into a mode-1 piece times a mode-2 piece:

\[ \braket{\tilde\psi}{\tilde\psi} = \bra{0}\hat a_1\hat a_1^\dagger\ket{0}\; \bra{0}\hat a_2^3(\hat a_2^\dagger)^3\ket{0} = 1\cdot \bra{0}\hat a_2^3(\hat a_2^\dagger)^3\ket{0}. \]

Now evaluate the remaining factor using repeated action on the vacuum. First note:

\[ (\hat a_2^\dagger)^3\ket{0}=\sqrt{3!}\,\ket{n_2=3}. \]

Applying \(\hat a_2^3\) brings \(\ket{n_2=3}\) back to \(\ket{0}\) with the same factorial factor:

\[ \hat a_2^3\ket{n_2=3}=\sqrt{3!}\,\ket{0}. \]

Therefore,

\[ \bra{0}\hat a_2^3(\hat a_2^\dagger)^3\ket{0} = \bra{0}\hat a_2^3\left(\sqrt{3!}\,\ket{n_2=3}\right) = \sqrt{3!}\,\bra{0}\left(\sqrt{3!}\,\ket{0}\right) = 3!. \]

So the norm is \(\braket{\tilde\psi}{\tilde\psi}=3!=6\), and the normalized state is

\[ \ket{\psi} = \frac{1}{\sqrt{3!}}\, (\hat a_1^\dagger)(\hat a_2^\dagger)^3\ket{0} = \frac{1}{\sqrt{6}}\,(\hat a_1^\dagger)(\hat a_2^\dagger)^3\ket{0}. \]

Equivalently, in occupation-number notation this is precisely the normalized Fock state with \((n_1,n_2)=(1,3)\):

\[ \ket{\psi}=\ket{n_1=1,n_2=3}. \]

(b) First-quantized wavefunction

Let the (normalized) symmetric \(4\)-boson wavefunction in the position basis be

\[ \Psi(x_1,x_2,x_3,x_4)=\braket{x_1,x_2,x_3,x_4}{\psi}. \]

Because the occupation numbers are \(n_1=1\) and \(n_2=3\), the wavefunction must be the fully symmetrized combination of “one particle in \(\phi_1\) and three particles in \(\phi_2\).”

A compact way to write it is with the symmetrizer \(\mathcal S\):

\[ \Psi(x_1,x_2,x_3,x_4) = \frac{1}{\sqrt{1!\,3!}}\; \mathcal S\Big[\phi_1(x_1)\phi_2(x_2)\phi_2(x_3)\phi_2(x_4)\Big]. \]

More explicitly (and very transparently), since there is exactly one \(\phi_1\), the symmetrized sum is just “choose which coordinate gets \(\phi_1\).” Define

\[ \Phi_j(x_1,x_2,x_3,x_4) \equiv \phi_1(x_j)\prod_{\ell\neq j}\phi_2(x_\ell). \]

Then the properly normalized bosonic wavefunction is

\[ \Psi(x_1,x_2,x_3,x_4) = \frac{1}{2}\sum_{j=1}^4 \Phi_j(x_1,x_2,x_3,x_4). \]

Writing the sum out:

\[ \Psi(x_1,x_2,x_3,x_4) = \frac{1}{2}\Big[ \phi_1(x_1)\phi_2(x_2)\phi_2(x_3)\phi_2(x_4) + \phi_2(x_1)\phi_1(x_2)\phi_2(x_3)\phi_2(x_4) \]

\[ + \phi_2(x_1)\phi_2(x_2)\phi_1(x_3)\phi_2(x_4) + \phi_2(x_1)\phi_2(x_2)\phi_2(x_3)\phi_1(x_4) \Big]. \]

Quick normalization check (why the prefactor is \(1/2\)): the four \(\Phi_j\) are mutually orthonormal because \(\phi_1\perp \phi_2\), so \(\|\sum_{j=1}^4 \Phi_j\|^2=4\), hence the normalized sum has coefficient \(1/\sqrt{4}=1/2\).

(c) Energy uncertainty of the many-body state

For noninteracting bosons in the harmonic oscillator, the second-quantized Hamiltonian is diagonal in the mode occupations:

\[ \hat H = \sum_{j} E_j\,\hat a_j^\dagger \hat a_j = \sum_j E_j\,\hat n_j. \]

Our state \(\ket{\psi}=\ket{n_1=1,n_2=3}\) is a simultaneous eigenstate of all number operators \(\hat n_j\), so it is an energy eigenstate:

\[ \hat H\ket{\psi} = (E_1 n_1 + E_2 n_2)\ket{\psi} = (E_1+3E_2)\ket{\psi}. \]

Therefore the total energy is definite and the energy uncertainty vanishes:

\[ \Delta E = 0. \]

(If you want the explicit value for the harmonic oscillator with \(\omega=\sqrt{k/m}\) and \(E_j=\hbar\omega(j-\tfrac{1}{2})\), then \(E_1=\tfrac{1}{2}\hbar\omega\) and \(E_2=\tfrac{3}{2}\hbar\omega\), so the total energy is \(E_{\text{tot}}=E_1+3E_2=5\hbar\omega\).)

(d) Energy of a single particle and its uncertainty

Even though the total energy is definite, the energy of a randomly chosen particle is not definite, because different bosons occupy different single-particle energy eigenstates.

Operationally: imagine measuring the single-particle energy of one boson chosen uniformly at random from the four bosons.

Because the occupation numbers are \(n_1=1\) and \(n_2=3\), the measurement outcomes are:

• With probability \(p_1=\frac{1}{4}\), the chosen boson is in orbital \(1\) and the outcome is \(E_1\).
• With probability \(p_2=\frac{3}{4}\), the chosen boson is in orbital \(2\) and the outcome is \(E_2\).

So the single-particle energy expectation value is

\[ \langle \varepsilon\rangle = \frac{1}{4}E_1+\frac{3}{4}E_2 = \frac{E_1+3E_2}{4}. \]

The variance is

\[ (\Delta \varepsilon)^2 = \frac{1}{4}(E_1-\langle \varepsilon\rangle)^2 + \frac{3}{4}(E_2-\langle \varepsilon\rangle)^2. \]

Compute the deviations from the mean:

\[ E_1-\langle \varepsilon\rangle = E_1-\frac{E_1+3E_2}{4} = \frac{3}{4}(E_1-E_2), \qquad E_2-\langle \varepsilon\rangle = E_2-\frac{E_1+3E_2}{4} = \frac{1}{4}(E_2-E_1). \]

Plugging in:

\[ (\Delta \varepsilon)^2 = \frac{1}{4}\cdot \frac{9}{16}(E_1-E_2)^2 + \frac{3}{4}\cdot \frac{1}{16}(E_2-E_1)^2 = \frac{3}{16}(E_2-E_1)^2. \]

Thus

\[ \Delta \varepsilon = \frac{\sqrt{3}}{4}\,|E_2-E_1|. \]

For the harmonic oscillator with \(E_1=\tfrac{1}{2}\hbar\omega\) and \(E_2=\tfrac{3}{2}\hbar\omega\), we have \(E_2-E_1=\hbar\omega\), so

\[ \Delta \varepsilon = \frac{\sqrt{3}}{4}\,\hbar\omega. \]

Conclusion: the energy of a single (randomly selected) particle is not definite; it is a two-valued random variable \(\varepsilon\in\{E_1,E_2\}\) with probabilities \(\{1/4,3/4\}\), and the uncertainty is as above.

Problem P-3: Two particle stimulated emission

Let \(N_{\text{tot}}=N+M_1+M_2\) be the total number of bosons.

It is convenient to interpret the Fock-state notation as

\[ \ket{N,M_1,M_2}\equiv \ket{n_{\vec k}=N,\; n_{\vec k-\vec q}=M_1,\; n_{\vec k+\vec q}=M_2}, \]

and similarly for the final state \(\ket{N-2,M_1+1,M_2+1}\).

Also expand the two-body potential as

\[ V_2(\vec R)=2V_0\cos(\vec q\cdot \vec R)=V_0\Big(e^{i\vec q\cdot \vec R}+e^{-i\vec q\cdot \vec R}\Big). \]

For a system of \(N_{\text{tot}}\) identical particles with coordinates \(\{\vec r_a\}\), the interaction operator is

\[ \hat V \;=\;\sum_{a<b} V_2(\vec r_a-\vec r_b). \]

The transition matrix element entering Fermi’s golden rule is \(\matrixel{f}{\hat V}{i}\), where \(\ket{i}=\ket{N,M_1,M_2}\) and \(\ket{f}=\ket{N-2,M_1+1,M_2+1}\) are properly normalized many-body states.

(a) Matrix element using first quantization (counting argument)

Step 1: Write normalized symmetrized many-body wavefunctions.

Let \(\phi_{\vec p}(\vec r)\) be the single-particle plane wave with momentum \(\vec p\) (any consistent normalization is fine; we will not evaluate integrals explicitly).

A convenient normalized bosonic wavefunction for the initial state is

\[ \Psi_i(\vec r_1,\dots,\vec r_{N_{\text{tot}}}) = \frac{1}{\sqrt{N!\,M_1!\,M_2!}}\; \mathcal S\Bigg[ \prod_{a=1}^{N}\phi_{\vec k}(\vec r_a)\, \prod_{b=N+1}^{N+M_1}\phi_{\vec k-\vec q}(\vec r_b)\, \prod_{c=N+M_1+1}^{N_{\text{tot}}}\phi_{\vec k+\vec q}(\vec r_c) \Bigg], \]

and the final state is

\[ \Psi_f(\vec r_1,\dots,\vec r_{N_{\text{tot}}}) = \frac{1}{\sqrt{(N-2)!\,(M_1+1)!\,(M_2+1)!}}\; \mathcal S\Bigg[ \prod_{a=1}^{N-2}\phi_{\vec k}(\vec r_a)\, \prod_{b=N-1}^{N+M_1-1}\phi_{\vec k-\vec q}(\vec r_b)\, \prod_{c=N+M_1}^{N_{\text{tot}}}\phi_{\vec k+\vec q}(\vec r_c) \Bigg]. \]

(Here \(\mathcal S\) denotes symmetrization over all particle labels.)

Step 2: Identify the piece of \(\hat V\) that connects the chosen occupations.

Using \(e^{\pm i\vec q\cdot(\vec r_a-\vec r_b)}=e^{\pm i\vec q\cdot \vec r_a}\,e^{\mp i\vec q\cdot \vec r_b}\), the interaction contains terms that act on a pair \((a,b)\) by multiplying one coordinate by \(e^{+i\vec q\cdot \vec r}\) and the other by \(e^{-i\vec q\cdot \vec r}\).

The key single-particle fact is:

\[ e^{\pm i\vec q\cdot \vec r}\,\phi_{\vec k}(\vec r)\;\propto\;\phi_{\vec k\pm \vec q}(\vec r), \]

so a term \(e^{+i\vec q\cdot \vec r_a}e^{-i\vec q\cdot \vec r_b}\) can convert a pair of \(\vec k\)-bosons into one \((\vec k+\vec q)\) boson and one \((\vec k-\vec q)\) boson (up to the same common spatial overlap integral for every such pair).

Thus, in \(\matrixel{f}{\hat V}{i}\), only those contributions survive where: - two bosons initially in mode \(\vec k\) are selected, - one is shifted to \(\vec k+\vec q\) and the other to \(\vec k-\vec q\).

Step 3: Do the combinatorics from symmetrization and normalization.

There are two independent “selection” effects:

1. Choosing the two \(\vec k\) bosons that participate.
   Because the initial state has \(N\) identical bosons in mode \(\vec k\), the number of ordered choices of two distinct \(\vec k\)-bosons is \(N(N-1)\). (This is the same physics as getting \(\sqrt{N(N-1)}\) when applying \(\hat a_{\vec k}\hat a_{\vec k}\) to a Fock state.)

2. Placing the created bosons into already occupied final modes.
   In the final state, the occupations are \(M_1+1\) in \((\vec k-\vec q)\) and \(M_2+1\) in \((\vec k+\vec q)\). When you take the overlap with the symmetrized final wavefunction, the number of equivalent placements contributes a factor that becomes \(\sqrt{(M_1+1)(M_2+1)}\) after accounting for the normalization of \(\Psi_f\) (this mirrors the \(\sqrt{M+1}\) factor from \(\hat a^\dagger\) acting on an occupied mode).

All spatial dependence factors into a single two-particle overlap integral (the same for every contributing choice). Define that common two-body matrix element by

\[ \tilde V_2(\vec q) \;\equiv\; \matrixel{\vec k-\vec q,\;\vec k+\vec q}{V_2(\vec r_1-\vec r_2)}{\vec k,\;\vec k}, \]

where the kets are properly symmetrized two-boson plane-wave states (this definition absorbs all convention-dependent volume factors).

Putting the counting and the common overlap together gives

\[ \boxed{ \matrixel{f}{\hat V}{i} = \tilde V_2(\vec q)\, \sqrt{N(N-1)(M_1+1)(M_2+1)} }. \]

Equivalently, the golden-rule rate is proportional to

\[ \boxed{ \left|\matrixel{f}{\hat V}{i}\right|^2 = \left|\tilde V_2(\vec q)\right|^2\, N(N-1)(M_1+1)(M_2+1)}. \]

This matches the structure in Aash’s handwritten solution. :contentReference[oaicite:0]{index=0}

(b) Matrix element using second quantization

Step 1: Write normalized initial and final Fock states.

Let \(\hat a_{\vec p}\) annihilate a boson in mode \(\vec p\). Then

\[ \ket{i} = \frac{(\hat a_{\vec k}^\dagger)^N(\hat a_{\vec k-\vec q}^\dagger)^{M_1}(\hat a_{\vec k+\vec q}^\dagger)^{M_2}}{\sqrt{N!\,M_1!\,M_2!}}\ket{0}, \qquad \ket{f} = \frac{(\hat a_{\vec k}^\dagger)^{N-2}(\hat a_{\vec k-\vec q}^\dagger)^{M_1+1}(\hat a_{\vec k+\vec q}^\dagger)^{M_2+1}}{\sqrt{(N-2)!\,(M_1+1)!\,(M_2+1)!}}\ket{0}. \]

Step 2: Express the relevant part of the interaction.

For a translationally invariant two-body potential, the interaction can be written in momentum space in terms of density operators

\[ \hat\rho_{\vec q} \equiv \sum_{\vec p}\hat a_{\vec p+\vec q}^\dagger \hat a_{\vec p}. \]

For the cosine potential, only the Fourier components at \(\pm \vec q\) appear, and the term that can connect \(\ket{i}\) to \(\ket{f}\) is proportional to \(\hat\rho_{\vec q}\hat\rho_{-\vec q}\). Focusing only on the piece that annihilates two \(\vec k\)-bosons and creates one \((\vec k-\vec q)\) and one \((\vec k+\vec q)\) boson, we may write

\[ \hat V \;\supset\; \tilde V_2(\vec q)\; \hat a_{\vec k-\vec q}^\dagger \hat a_{\vec k+\vec q}^\dagger \hat a_{\vec k}\hat a_{\vec k}, \]

where (as in part (a)) \(\tilde V_2(\vec q)\) denotes the appropriate two-body Fourier matrix element and absorbs any convention-dependent prefactors.

Step 3: Apply ladder operators step-by-step.

Act on the initial state:

\[ \hat a_{\vec k}\hat a_{\vec k}\ket{i} = \sqrt{N}\sqrt{N-1}\; \ket{n_{\vec k}=N-2,\;n_{\vec k-\vec q}=M_1,\;n_{\vec k+\vec q}=M_2}. \]

Then create into the other two modes:

\[ \hat a_{\vec k-\vec q}^\dagger \hat a_{\vec k+\vec q}^\dagger \ket{n_{\vec k}=N-2,\;n_{\vec k-\vec q}=M_1,\;n_{\vec k+\vec q}=M_2} = \sqrt{M_1+1}\sqrt{M_2+1}\;\ket{f}. \]

Therefore,

\[ \matrixel{f}{\hat V}{i} = \tilde V_2(\vec q)\, \sqrt{N(N-1)}\sqrt{(M_1+1)(M_2+1)}. \]

So we again obtain

\[ \boxed{ \matrixel{f}{\hat V}{i} = \tilde V_2(\vec q)\, \sqrt{N(N-1)(M_1+1)(M_2+1)} }. \]

(c) Heuristic explanation of each factor

From Fermi’s golden rule,

\[ \Gamma_{i\to f}\propto \left|\matrixel{f}{\hat V}{i}\right|^2 = \left|\tilde V_2(\vec q)\right|^2\,N(N-1)(M_1+1)(M_2+1). \]

Each factor has a clear origin:

• \(\left|\tilde V_2(\vec q)\right|^2\) is the squared strength of the interaction at the required momentum transfer \(\pm \vec q\) (the relevant Fourier component of the two-body potential).

• \(N(N-1)\) reflects the availability of two bosons in the initial \(\vec k\) mode to participate in the process. In operator language, annihilating two bosons gives \(\hat a_{\vec k}^2\ket{N}\propto \sqrt{N(N-1)}\ket{N-2}\), and squaring the amplitude gives \(N(N-1)\).

• \((M_1+1)(M_2+1)\) is the bosonic stimulation factor for emission into already occupied final modes \((\vec k-\vec q)\) and \((\vec k+\vec q)\). Creating in an occupied bosonic mode gives \(\hat a^\dagger\ket{M}\propto \sqrt{M+1}\ket{M+1}\), and the rate depends on \((M+1)\) for each created mode.

• The “\(+1\)” terms encode the fact that even if \(M_1=0\) or \(M_2=0\), the transition can still occur (the analogue of spontaneous emission). When \(M_1\) and/or \(M_2\) are large, the rate is strongly enhanced due to stimulated emission.